SECTION 19.3. Slightly Soluble Ionic Compounds a Very Soluble Ionic Salt (E.G., Nacl) Dissolves and Completely Dissociates Into Na+ (Aq) and Cl- (Aq)

SECTION 19.3. Slightly Soluble Ionic Compounds A very soluble ionic salt (e.g., NaCl) dissolves and completely dissociates into Na+ (aq) and Cl- (aq). However, some salts are only slightly soluble, and an equilibrium exists between dissolved and undissolved compound. Consider the addition of

PbSO4 (s) to water. 2+ 2- PbSO4 (s) ⇌ Pb (aq) + SO4 (aq)

If the reaction hasn’t reached equilibrium, the reaction quotient Qc is [Pb2 ][SO 2] Qc = 4 [PbSO4]

The concentration of a solid (= its density) is a constant  combine it with Qc. 2+ 2- Qc [PbSO4] = Qsp = [Pb ][SO4 ]

Qsp = “ion-product expression”. At equilibrium, Qsp =Ksp 2+ 2- Ksp =[Pb ][SO4 ]

Ksp = “Solubility Product Constant” or just "Solubility Product"

Ksp, like other equilibrium constants, only depends on temperature.

n+ z- For any salt MpXq (s) ⇌ pM (aq) + qX (aq)

n+ P z- q Ksp = [M ] [X ] 19-1 Examples:

2+ - 2+ - 2 Cu(OH)2 (s) ⇌ Cu (aq) + 2 OH (aq) Ksp =[Cu ][OH ] 2+ 2- 2+ 2- CaCO3 (s) ⇌ Ca (aq) + CO3 (aq) Ksp =[Ca ][CO3 ] 2+ 3- 2+ 3 3- 2 Ca3(PO4)2 (s) ⇌ 3Ca (aq) + 2 PO4 (aq) Ksp =[Ca ] [PO4 ]

Special case of metal sulfides: The S2- (aq) anion, a weak base (but one of the strongest weak bases) is very unstable in water and undergoes (~100%) a base- dissociation reaction to give HS- (aq) and OH- (aq).

2- - - S (aq) + H2O(l) HS (aq) + OH (aq)  the solubility equilibrium when MnS (s) is dissolved in water is

2+ - - MnS (s) + H2O(l)⇌ Mn (aq) + HS (aq) + OH (aq) 2+ - -  Ksp =[Mn ][HS ][OH ] ------

** Now, the greater is Ksp, the more soluble the substance is. ** -8 e.g., PbSO4 Ksp =1.6x10 insoluble -10 CoCO3 Ksp =1.0x10 more insoluble (or less soluble) -15 Fe(OH)2 Ksp =4.1x10 most insoluble (or least soluble) Table 19.2

19-2 Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic Compounds at 25°C

Name, Formula Ksp

-34 Aluminum hydroxide, Al(OH)3 3x10

-10 Cobalt(II) carbonate, CoCO3 1.0x10

-15 Iron(II) hydroxide, Fe(OH)2 4.1x10

-8 Lead(II) fluoride, PbF2 3.6x10

-8 Lead(II) sulfate, PbSO4 1.6x10

-29 Mercury(I) iodide, Hg2I2 4.7x10

-48 Silver sulfide, Ag2S 8x10

-6 Zinc iodate, Zn(IO3)2 3.9x10

19-3 Table 19.3 Relationship Between Ksp and Solubility at 25°C

No. of Ions Formula Cation/Anion Ksp Solubility (M) -8 -4 2MgCO3 1/1 3.5x10 1.9x10 -8 -4 2PbSO4 1/1 1.6x10 1.3x10 -10 -5 2BaCrO4 1/1 2.1x10 1.4x10

-6 -2 3Ca(OH)2 1/2 6.5x10 1.2x10 -6 -3 3BaF2 1/2 1.5x10 7.2x10 -11 -4 3CaF2 1/2 3.2x10 2.0x10 -12 -5 3Ag2CrO4 2/1 2.6x10 8.7x10

** The higher the Ksp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas. **

19-4 Calculations Involving Solubility Products

Two types: Use Ksp to find concs of dissolved ions Use concs to find Ksp.

Example: The solubility of Ag2CO3 is 0.032 M at 20 °C. What is its Ksp? A common type of question - note that we are told the molar solubility of

Ag2CO3, but of course it will dissociate into ions. Therefore, + 2- Ag2CO3 (s)  2 Ag (aq) + CO3 (aq) [init] (solid) 0 0 [change] -0.032 M +0.064 +0.032 [equil] (solid) 0.064 M 0.032 M

+ 2 2 Ksp = [Ag ] [CO3] = (0.064) (0.032)

-4 Ksp = 1.3 x 10

-3 Example: The solubility of Zn (oxalate) is 7.9 x 10 M at 18 °C. What is its Ksp? Zn (ox) ⇌ Zn2+ (aq) + ox2- (aq) [init] (solid) - [change] -7.9 x 10-3 M +7.9 x 10-3 +7.9 x 10-3 [equil] (solid) 7.9 x 10-3 M7.9 x 10-3 M

2+ -3 2 -5 Ksp = [Zn ][ox] = (7.9 x 10 ) Ksp = 6.2 x 10

19-5 -10 Example: What is the molar solubility of SrCO3? (Ksp = 5.4 x 10 ) 2+ 2- SrCO3 (s) ⇌ Sr (aq) + CO3 (aq) [init] (solid) 0 0 change -x +x +x [equil] (solid) x x

2 -5 Ksp = x  x =  x = 2.3 x 10 M

-5  Solubility of SrCO3 is 2.3 x 10 M

-6 Example: What is the molar solubility of Ca(OH)2? (Ksp = 6.5 x 10 ). 2+ - Ca(OH)2 (s) ⇌ Ca (aq) + 2 OH (aq) [init] (solid) 0 0 change -x +x +2x [equil] (solid) x 2x

-6 2 3 Ksp = 6.5 x 10 = x(2x) = 4x (careful!) -2  Solubility of Ca(OH)2 = 1.2 x 10 M

To obtain the x of a number, learn to use the x button on your calculator, or take the log, divide by x, then antilog.

19-6 The Effect of a Common Ion on Solubility Addition of a common ion decreases solubility - due to Le Chatelier’s principle.

2+ 2- PbCrO4 (s) ⇌ Pb (aq) + CrO4 (aq) 2+ 2- -13 Ksp = [Pb ][CrO4 ] = 2.3 x 10

Consider this PbCrO4 system at equilibrium. If we dissolve some Na2CrO4 (s) (very 2- soluble) in the solution, what will happen? [CrO4 ] will increase, because the 2- Na2CrO4 will dissociate to give more CrO4 . Na CrO (s) 2 Na+ (aq) + CrO 2- (aq) 2 4 100%  4 2+ 2- 2- The Ksp for PbCrO4 tells us that [Pb ][CrO4 ] is a constant  if [CrO4 ]increases, 2+ [Pb ] must decrease. How can that happen? Some PbCrO4 (s) precipitates from solution, i.e., the equilibrium shifts to the left.

2+ 2- PbCrO4 (s) ⇌ Pb (aq) + CrO4 (aq) 2- add CrO4 Shift

The equilibrium will shift to the left (i.e., more PbCrO4 (s) will form) until 2+ 2- -13 [Pb ]new[CrO4 ]new =Ksp =2.3x10 .

** i.e. PbCrO4 is less soluble in aqueous Na2CrO4 solution than in pure water. ** 2+ Similarly, addition of a soluble source of Pb (e.g., Pb(NO3)2) to a solution of 2+ PbCrO4 will increase [Pb ]andcausePbCrO4 (s) to precipitate from solution. Fig 19.12 19-7 Figure 19.12 The effect of a common ion on solubility.

2+ 2- PbCrO4(s) Pb (aq) + CrO4 (aq)

If Na2CrO4 solution is added to a saturated solution of PbCrO4, it 2- provides the common ion CrO4 , causing the equilibrium to shift to the left. Solubility decreases and solid PbCrO4 precipitates.

19-8 To understand this, let’s make up an example that will use more convenient numbers.

Imagine a solid MX with Ksp = 100. MX (s) ⇌ M+ (aq) + X- (aq) + - at equilibrium, [M ] = 10.0 M [X ] = 10.0 M  Ksp = 10.0x10.0 = 100 + - + - (i.e MX(s) will dissolve until [M ]=[X ]=10.0MbecauseKsp =[M][X ]=100) Now, let’s add enough very soluble NaX to make new [X-]=20M,i.e., we double [X-]. + - now [M ][X ] = 10.0x20.0 = 200 = Qsp  Ksp. Reaction no longer at equilibrium  shifts to the left until it returns to equilibrium  more MX (s) forms, decreasing [M+]and[X-].

+ - When [M ]=10M [X] = 20M Qsp = 200  Ksp + - [M ]=9M [X]=19M Qsp =171 Ksp + - [M ]=8M [X]=18M Qsp =144 Ksp + - [M ]=7M [X]=17M Qsp = 119  Ksp + - [M ] = 6.5M [X ]=16.5M Qsp =107 Ksp + - [M ] = 6.2M [X ]=16.2M Qsp =100=Ksp  reaction at equilibrium again with new [M+]=6.2M,new[X-]=16.2M Compare with old [M+]=10.0M,new[X-]=10.0M

+ - Qsp =[M][X ] = (6.2)(16.2) = 100 = Ksp.

Remember: Le Chatelier’s principle tells us that if we perturb a reaction at equilibrium, it will shift in a direction that allows it to reach equilibrium again. 19-9 Effect of pH on Solubility

Similar to the previous section, if a compound contains the anion of a + weak acid, addition of H3O (from a strong acid) increases its solubility.

Why? Le Chatelier’s principle again.

2+ 2- CaCO3 (s) ⇌ Ca (aq) + CO3 (aq)

+ 2- Addition of H3O causes the [CO3 ] to decrease.

2- + - CO3 (aq) + H3O (aq)  HCO3 (aq) + H2O

2+ 2- CaCO3 (s) ⇌ Ca (aq) + CO3 (aq) 2- if [CO3 ] decreases Shift 2- + i.e., when we decrease [CO3 ] (by adding H3O ), reaction will shift to 2- the right to make more CO3 and restore equilibrium.

19-10 Figure 19.13 Test for the presence of a carbonate.

When a carbonate mineral is treated with HCl, bubbles of CO2 form - + HCO3 (aq) + H3O (aq)  H2CO3 (aq) + H2O

H2CO3 (aq)  CO2 (g) + H2O.

19-11 Predicting the Formation of a Precipitate: Qsp =Ksp

Three possibilities:

(a) If Qsp =Ksp, reaction at equilibrium, no change.

(b) If Qsp > Ksp, reaction not at equilibrium

 [ ] of ions in solution too high

 solid (precipitate) forms until Qsp =Ksp

 [ ] of ions in solution too low

 no solid (precipitate) forms, i.e. nothing happens

19-12 Example: Will a precipitate form if 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

Method: Calculate Qsp and compare with Ksp. + - First: What is the sparingly soluble salt? NaNO3,likeallNa and NO3 salts, is very soluble in water  it must be CaF2. 2+ -  CaF2 (s) ⇌ Ca (aq) + 2 F (aq) 2+ - 2 2+ - Ksp =[Ca ][F ]  need [Ca ]and[F].

2+ (a) We start with 0.30 M Ca(NO3)2 =0.30M[Ca ] (~100% dissociation) (b) We start with 0.060 M NaF = 0.060 M [F-] (~100% dissociation)

-- but we are mixing two solutions, therefore both will be diluted.

Remember: MiVi =MfVf (i = initial, f = final) 2 2+ [Ca ]iVi (0.30 M)(0.100 L) 2+  [Ca ]f ==  [Ca ]f =0.10M Vf (0.200 L  0.100 L) - - [F ]iVi (0.060 M)(0.200 L) - [F ]f ==  [F ]f = 0.040 M Vf (0.300 L) 2+ - 2 2 -4 Qsp =[Ca ][F ] = (0.10)(0.040) =1.6x10 -11 Ksp =3.2x10  Qsp >Ksp  CaF2 (s) precipitates until Qsp =Ksp 19-13 Section 19.4 Solubility Equilibria involving Complex Ions AgCl (s) is only slightly soluble, and it does NOT become more soluble if we lower the pH (because Cl- is the conjugate base of a strong acid and therefore does not bind the added H+)

BUT, AgCl (s) becomes much more soluble if we add NH3 to the solution. Why? - + because the NH3 reacts with the Ag (aq) (Lewis acid/Lewis base reaction) to give + acomplexion[Ag(NH3)2] . Consider what happens as two steps: + - -10 AgCl (s) ⇌ Ag (aq) + Cl (aq) Ksp =1.8x10 (v. small) + + 7 Ag (aq) + 2 NH3 (aq) ⇌ [Ag(NH3)2] K=Kform =1.6x10 ** The equilibrium constant of a reaction which is the formation of a complex

ion is called the “formation constant”, Kform or Kf **

So, the overall reaction that occurs when AgCl is dissolved in water with NH3 also present is + - AgCl (s) + 2 NH3 (aq) ⇌ [Ag(NH3)2] +Cl (aq) (Knet)

The net Kc for this (Knet) is the product of the Ksp and Kform i.e. the K’s of the individual two steps.

-10 7 -3  Knet =Ksp xKf = (1.8 x 10 )(1.6 x 10 )= 2.9x10  therefore the solubility of the AgCl (s) is indeed much greater when

NH3 is also present, since Knet >> Ksp (by 10 million times !) 19-14 3+ Figure 19.15 Cr(NH3)6 , a typical complex ion.

A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands.

19-15 Table 19.4 Formation Constants (Kf) of Some Complex Ions at 25°C